Question

How do you solve the system of equations?
$\begin{aligned} & 3x+2y+4z=11 \\\ & 2x-y+3z=4 \\\ & 5x-3y+5z=-1 \\\ \end{aligned}$

Answer 1

The given system of equations can be solved by using the Cramer’s rule. For this, we first have to express the equations in the matrix form. Then we need to calculate the determinant, $\Delta$ of the coefficient matrix. Then, we have to calculate the determinants, ${{\Delta }_{1}},{{\Delta }_{2}},{{\Delta }_{3}}$ of the matrices obtained by replacing the first, second and the third columns of the coefficient matrix respectively. The final solution will be obtained as $x=\dfrac{{{\Delta }_{1}}}{\Delta },y=\dfrac{{{\Delta }_{2}}}{\Delta },z=\dfrac{{{\Delta }_{3}}}{\Delta }$.

Complete step by step solution:
The given system of equations is
$\begin{aligned} & 3x+2y+4z=11 \\\ & 2x-y+3z=4 \\\ & 5x-3y+5z=-1 \\\ \end{aligned}$
The above system of the equations can be expressed in terms of matrices as
$\Rightarrow \left[ \begin{matrix} 3 & 2 & 4 \\\ 2 & -1 & 3 \\\ 5 & -3 & 5 \\\ \end{matrix} \right]\left[ \begin{matrix} x \\\ y \\\ z \\\ \end{matrix} \right]=\left[ \begin{matrix} 11 \\\ 4 \\\ -1 \\\ \end{matrix} \right]$
From above, the we calculate the determinant of the coefficient matrix as

& \Rightarrow \Delta =\left| \begin{matrix} 3 & 2 & 4 \\\ 2 & -1 & 3 \\\ 5 & -3 & 5 \\\ \end{matrix} \right| \\\ & \Rightarrow \Delta =3\left[ \left( -1 \right)\left( 5 \right)-\left( -3 \right)\left( 3 \right) \right]-2\left[ \left( 2 \right)\left( 5 \right)-\left( 5 \right)\left( 3 \right) \right]+4\left[ \left( 2 \right)\left( -3 \right)-\left( 5 \right)\left( -1 \right) \right] \\\ & \Rightarrow \Delta =3\left( -5+9 \right)-2\left( 10-15 \right)+4\left( -6+5 \right) \\\ & \Rightarrow \Delta =3\left( 4 \right)-2\left( -5 \right)+4\left( -1 \right) \\\ & \Rightarrow \Delta =12+10-4 \\\ & \Rightarrow \Delta =18........\left( i \right) \\\ \end{aligned}$$ Thus, the determinant of the coefficient matrix is non-zero. Now, we replace the first column of the coefficient matrix by the constant terms and calculate the determinant of the resultant matrix as $$\begin{aligned} & \Rightarrow {{\Delta }_{1}}=\left| \begin{matrix} 11 & 2 & 4 \\\ 4 & -1 & 3 \\\ -1 & -3 & 5 \\\ \end{matrix} \right| \\\ & \Rightarrow {{\Delta }_{1}}=11\left[ \left( -1 \right)\left( 5 \right)-\left( -3 \right)\left( 3 \right) \right]-2\left[ \left( 4 \right)\left( 5 \right)-\left( -1 \right)\left( 3 \right) \right]+4\left[ \left( 4 \right)\left( -3 \right)-\left( -1 \right)\left( -1 \right) \right] \\\ & \Rightarrow {{\Delta }_{1}}=11\left[ -5+9 \right]-2\left[ 20+3 \right]+4\left[ -12-1 \right] \\\ & \Rightarrow {{\Delta }_{1}}=11\left( 4 \right)-2\left( 23 \right)+4\left( -13 \right) \\\ & \Rightarrow {{\Delta }_{1}}=44-46-52 \\\ & \Rightarrow {{\Delta }_{1}}=-54........\left( ii \right) \\\ \end{aligned}$$ Now, we replace the second column of the coefficient matrix by the constant matrix and determine its determinant as $$\begin{aligned} & \Rightarrow {{\Delta }_{2}}=\left| \begin{matrix} 3 & 11 & 4 \\\ 2 & 4 & 3 \\\ 5 & -1 & 5 \\\ \end{matrix} \right| \\\ & \Rightarrow {{\Delta }_{2}}=3\left[ \left( 4 \right)\left( 5 \right)-\left( -1 \right)\left( 3 \right) \right]-11\left[ \left( 2 \right)\left( 5 \right)-\left( 5 \right)\left( 3 \right) \right]+4\left[ \left( 2 \right)\left( -1 \right)-\left( 5 \right)\left( 4 \right) \right] \\\ & \Rightarrow {{\Delta }_{2}}=3\left[ 20+3 \right]-11\left[ 10-15 \right]+4\left[ -2-20 \right] \\\ & \Rightarrow {{\Delta }_{2}}=3\left( 23 \right)-11\left( -5 \right)+4\left( -22 \right) \\\ & \Rightarrow {{\Delta }_{2}}=69+55-88 \\\ & \Rightarrow {{\Delta }_{2}}=36........\left( iii \right) \\\ \end{aligned}$$ Similarly, we replace the third column and calculate the determinant as $$\begin{aligned} & \Rightarrow {{\Delta }_{3}}=\left| \begin{matrix} 3 & 2 & 11 \\\ 2 & -1 & 4 \\\ 5 & -3 & -1 \\\ \end{matrix} \right| \\\ & \Rightarrow {{\Delta }_{3}}=3\left[ \left( -1 \right)\left( -1 \right)-\left( -3 \right)\left( 4 \right) \right]-2\left[ \left( 2 \right)\left( -1 \right)-\left( 5 \right)\left( 4 \right) \right]+11\left[ \left( 2 \right)\left( -3 \right)-\left( 5 \right)\left( -1 \right) \right] \\\ & \Rightarrow {{\Delta }_{3}}=3\left[ 1+12 \right]-2\left[ -2-20 \right]+11\left[ -6+5 \right] \\\ & \Rightarrow {{\Delta }_{3}}=3\left( 13 \right)-2\left( -22 \right)+11\left( -1 \right) \\\ & \Rightarrow {{\Delta }_{3}}=39+44-11 \\\ & \Rightarrow {{\Delta }_{3}}=72........\left( iv \right) \\\ \end{aligned}$$ Now, from the Cramer’s rule we know that the solution is given by $\Rightarrow x=\dfrac{{{\Delta }_{1}}}{\Delta },y=\dfrac{{{\Delta }_{2}}}{\Delta },z=\dfrac{{{\Delta }_{3}}}{\Delta }$ Substituting (i), (ii), (iii) and (iv) we get $\begin{aligned} & \Rightarrow x=\dfrac{-54}{18},y=\dfrac{36}{18},z=\dfrac{72}{18} \\\ & \Rightarrow x=-3,y=2,z=4 \\\ \end{aligned}$ **Hence, the solution of the given system of equations is $x=-3,y=2,z=4$.** **Note:** The above question is solved by using the Cramer’s rule. But we can also solve this question by using the substitution method. For that we need to obtain the value of any variable from the first equation in terms of the other two variables, and put it into the other two equations to get two equations in two variables. Solving the two equations, we will get their values, which on substituting into any one equation, we will get the required solution.